如何用Python实现LeetCode241.不同加括号方式数问题？
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本文共计298个文字，预计阅读时间需要2分钟。
题目：https://leetcode.com/problems/different-ways-to-add-parentheses/description/+给定一个由数字和运算符组成的字符串，返回所有可能的计算结果，通过计算所有可能的数字和运算符组合方式来分组数字。
+示例：+输入：(2+3)*(5+6)+输出：[23, 52]
题目：leetcode.com/problems/different-ways-to-add-parentheses/description/
Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.
Example 1:
Input: “2-1-1”.
((2-1)-1) = 0(2-(1-1)) = 2
Output: [0, 2]
Example 2:
Input: “23-45”
(2*(3-(45))) = -34((23)-(45)) = -14((2(3-4))5) = -10(2((3-4)5)) = -10(((23)-4)*5) = 10
Output: [-34, -14, -10, -10, 10]
分析：递归，笛卡尔积代码：
class Solution: def diffWaysToCompute(self, input): ops = {'+': lambda x, y: x + y, '-': lambda x, y: x - y, '*': lambda x, y: x * y} def ways(s): ans = [] for i in range(len(s)): if s[i] in "+-*": ans += [ops[s[i]](l, r) for l, r in itertools.product(ways(s[0:i]), ways(s[i+1:]))] if not ans: ans.append(int(s)) return ans return ways(input)代码转自：zxi.mytechroad.com/blog/leetcode/leetcode-241-different-ways-to-add-parentheses/
本文共计298个文字，预计阅读时间需要2分钟。
题目：https://leetcode.com/problems/different-ways-to-add-parentheses/description/+给定一个由数字和运算符组成的字符串，返回所有可能的计算结果，通过计算所有可能的数字和运算符组合方式来分组数字。
+示例：+输入：(2+3)*(5+6)+输出：[23, 52]
题目：leetcode.com/problems/different-ways-to-add-parentheses/description/
Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.
Example 1:
Input: “2-1-1”.
((2-1)-1) = 0(2-(1-1)) = 2
Output: [0, 2]
Example 2:
Input: “23-45”
(2*(3-(45))) = -34((23)-(45)) = -14((2(3-4))5) = -10(2((3-4)5)) = -10(((23)-4)*5) = 10
Output: [-34, -14, -10, -10, 10]
分析：递归，笛卡尔积代码：
class Solution: def diffWaysToCompute(self, input): ops = {'+': lambda x, y: x + y, '-': lambda x, y: x - y, '*': lambda x, y: x * y} def ways(s): ans = [] for i in range(len(s)): if s[i] in "+-*": ans += [ops[s[i]](l, r) for l, r in itertools.product(ways(s[0:i]), ways(s[i+1:]))] if not ans: ans.append(int(s)) return ans return ways(input)代码转自：zxi.mytechroad.com/blog/leetcode/leetcode-241-different-ways-to-add-parentheses/